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The $j$th column of the matrix representation of a linear map is the coordinate vector of the image of the $j$th basis vector:Define $C$ by $w_j={\sum }_i{C}_{ij}{u}_i$, so its columns express the new basis vectors in the old basis. For every vector, $[v{]}_u=C[v{]}_w$. Applying $\varphi$ in either coordinate system gives $AC[v{]}_w=C{A}^{\prime }[v{]}_w$, proving the change of basis formulaTo make $A^{\prime }$ diagonal, choose the columns of $C$ to be linearly independent eigenvectors of $A$, expressed in the $u$ basis, and put their corresponding eigenvalues on the diagonal in the same order. This is possible precisely when $A$ is a diagonalizable matrix over the stated field; an eigenbasis need not exist over an arbitrary field.

Split the vector $x$ into its normal and tangential orthogonal projections:A reflection in the plane negates the first component and preserves the second. Thus $Hx=x-2(n\cdot x)n$, giving the Householder reflectionSince $n\cdot n=1$ and $n\cdot u=n\cdot v=0$,The columns of a matrix representation of a linear map are the coordinates of the images of the ordered basis vectors. Consequently, in the orthonormal basis $(u,v,n)$,Equivalently, with the orthogonal matrix $Q=(uvn)$, the change of basis formula is $N=Q^{T}MQ$.

Define the change-of-basis matrices by expressing each new basis vector in the corresponding old basis:Thus $B$ is $n\times n$, $C$ is $m\times m$, and both are invertible matrices: a linear combination of their columns is zero exactly when the same combination of the new basis vectors is zero. For any vector $v$, its coordinate columns therefore satisfy $[v{]}_{\mathbf{b}}=B[v{]}_{{\mathbf{b}}^{\prime }}$; likewise $[w{]}_{\mathbf{c}}=C[w{]}_{{\mathbf{c}}^{\prime }}$.

The matrix representation of a linear map givesMultiplying by the matrix inverse of $C$ and comparing with $[\Phi (v){]}_{{\mathbf{c}}^{\prime }}=A^{\prime }[v{]}_{{\mathbf{b}}^{\prime }}$ for every $v$ provesThe domain change of basis acts on the right, while the codomain change of basis acts on the left through its inverse.

Let $P_B,P_{B^{\prime }},P_C,P_{C^{\prime }}$ have the corresponding basis vectors as columns in standard coordinates. From the PDF,The matrix representation of a linear map in standard coordinates is $P_{C}A{P}_B^{-1}$. The change of basis formula therefore givesMultiplication yields

For ordered bases $\mathcal{B}=(v_1,\dots ,v_n)$ of $V$ and $\mathcal{C}=(w_1,\dots ,w_m)$ of $W$, the matrix representation of a linear map $\gamma :V\to W$ is the matrix whose $j$th column consists of the $\mathcal{C}$-coordinates of $\gamma (v_j)$. Thus

The operators $\alpha ,\beta :V\to V$ are conjugate linear operators when there is an isomorphism $s:V\to V$ such thatIn one basis their matrices therefore satisfy $[\beta ]=[s{]}^{-1}[\alpha ][s]$, so they are similar matrices.

For invertible $\beta$, the mapis linear, and its inverse is $A\mapsto \beta A{\beta }^{-1}$; hence it is a linear isomorphism of $\mathrm{End}(V)$. If ${\beta }^{\prime }=s^{-1}\beta s$, define ${\Phi }_s(A)=s^{-1}As$. A direct substitution givesso ${\varphi }_{{\beta }^{\prime }}$ and ${\varphi }_{\beta }$ are conjugate. This is the conjugation operator on an endomorphism space.

It remains to compute the Jordan normal form over $\mathbb{C}$. By the preceding conjugacy, we may put $\beta$ in Jordan form.

Ifthen the matrix units $E_{ij}$ are eigenvectors becauseThusThis includes the scalar case $\lambda =\mu$, when ${\varphi }_{\beta }$ is the identity.

Otherwise $\beta$ has one size-two Jordan block. Multiplying $\beta$ by a nonzero scalar does not change ${\varphi }_{\beta }$, and conjugating within its Jordan class allows us to use $J=I+N$ with $N=E_{12}$. Put $D={\varphi }_J-I$. Direct multiplication givesHence $D^3=0$, $\mathrm{rank}D=2$, and $\mathrm{rank}{D}^2=1$. The nilpotent Jordan blocks of $D$ therefore have sizes three and one. Adding the identity givesThese two cases are the Jordan normal form of conjugation on two-by-two matrices.

$A$ is not an orthogonal matrix. Its first column has squared Euclidean normso $A^{T}A\ne I$. The reason this is compatible with ii is that $1,x,x^2$ is not an orthonormal basis for the given inner product. Its Gram matrix isThe correct condition on this matrix representation of a linear map iswhich follows either by direct multiplication or by the inner product preservation just proved. This is the rule for an orthogonal transformation in a nonorthonormal basis.

Choose the Fourier series conventionFor a real potential energy, $V_{-1}=V_1^{\ast }$. The symbol $V(G_1)$ means the Fourier coefficient $V_1$, with units of energy, rather than the pointwise value of $V(x)$ at a position $G_1$.

Retain the two plane waves $|1⟩=L^{-1/2}{e}^{ikx}$ and $|2⟩=L^{-1/2}{e}^{i(k-G_1)x}$ in the nearly-free electron model. Their kinetic energies are ${ϵ}_1={\hslash }^2{k}^2/(2m)$ and ${ϵ}_2={\hslash }^2(k-G_1{)}^2/(2m)$. The matrix representation of a linear map for the Quantum Hamiltonian projected onto this two-dimensional basis isThus $\det (H-EI)=0$ givesSolving this quadratic equation yieldsChoosing the zero of potential energy so that its spatial average is $V_0=0$ reproduces the requested expression. If that average has not been removed, both energy bands are shifted by $V_0$; no band gap, group velocity or effective mass is changed by this constant shift.

The two kinetic energies coincide at $k=G_1/2$. Their coupling gives an avoided crossing with band gap $2|V_1|$. The result is exact within the projected two-wave Quantum Hamiltonian. It is an approximation to the full crystal problem, appropriate near this crossing for a weak periodic potential whose other Fourier coefficients do not cause strong mixing with remote plane waves. Small $|V_1|$ alone would not justify ignoring arbitrarily large other Fourier coefficients.